Algebra Basics: Slope And Distance – Math Antics

Algebra Basics: Slope And Distance – Math Antics


Hi, I’m Rob. Welcome to Math Antics! In our video about Basic Linear Functions, we learned that the equation y=mx + b can be used to represent any linear function on the 2D coordinate plane. In this video, we’re going to dive a little deeper and learn two new equations: one for calculating the slope of any line, and the other for calculating the distance between any two points on a line. To do that, the first thing we need is two points. Yeah! Two points!! Uh actually, I meant two points on the coordinate plane, but nice shot anyway. We’re gonna name these points “Point 1” and “Point 2” As you know from Geometry, a line can be defined by any two points. All you need to do is “connect the dots” to get a line segment. And you can get the infinite line by extending that line segment in either direction. Our goal in this video is to learn how to find the slope of a line segment like this and to calculate the distance between its two endpoints. The key to accomplishing those goals is to realize that you can use any line segment to make a right triangle. To do that, start from the point that is the highest on the coordinate plane and draw a vertical line straight down towards the bottom of the coordinate plane. Next, move to the other point and draw a horizontal line until it intersects the vertical line you just drew. See… now you have a right triangle. The vertical and horizontal sides of the right triangle are perpendicular to each other so they form a right angle. And the original diagonal line segment is now the hypotenuse of the new triangle. Temporarily, we’re going to call the horizontal side of the triangle “change in x” and the vertical side “change in y”, which are awkwardly long names, but that’s exactly what the two sides represent. Imagine starting at Point 1 and then slowly moving along our line segment towards Point 2. As you do that, the X and Y coordinates that you’re located at are changing, right? Your X- coordinate is changing because you’re traveling to the right, and your Y-coordinate is changing because you’re traveling up. And when you finally get to Point 2, the total change in your X-coordinate would be the length of the horizontal side of the triangle, and the total change in your Y-coordinate would be the length of the vertical side of the triangle. That’s cool, but the names “change in X” and ”change in Y” are kinda long. Fortunately, mathematicians have a shorter way of saying the same thing. They use the greek letter “delta” as an abbreviation for the words “change in”. That means you can just write or say “delta X” to mean the “change in X” and “delta Y” to mean the “change in Y”. Okay, now that we know what those sides represent and we’ve got nice names for them, how do we actually calculate the lengths of those sides? To do that, we just need the coordinates of the two points that form our line. Since we named them Point 1 and Point 2, it makes sense for us to call their coordinate values “X1 and Y1” for Point 1 and “X2 and Y2” for Point 2. We write the 1s and 2s as subscripts after the variables so we don’t mistakenly think it means X times 1 or X times 2. The subscript numbers are simply a way of distinguishing the different variables in the problem so we can keep track of which is which. Anyway, to calculate the (delta X), you need to find the difference between the X-coordinates of the two points. In other words, you need to subtract the X-value of the 1st point from the X-value of the 2nd point. So (delta X)=X2 − X1 Likewise, to calculate (delta Y), you need to find the difference between the Y-coordinates of the two points. So (delta Y)=Y2 − Y1 These simple equations for finding (delta X) and (delta Y) are important because they’re used in the equations for slope and distance, which is what I want to show you now. The equation for the slope of a line looks like this: Slope=(delta Y) over (delta X). Remember that the fraction line means division, so (delta Y) over (delta X) is the same as (delta Y) divided by (delta X). You might also see the slope equation written in expanded form like this. The only difference is that the (delta X) and (delta Y) have been written out to show the subtractions you need to do to get those values from the coordinates. Either way you want to write it is fine. And some of you may have heard this same equation expressed using different terminology. Have you ever heard someone say that “slope=rise over run”? “Rise” and “Run” are just different names that are sometimes used to describe the change in Y and the change in X. The idea is that if you were to “run” along the line in the X direction, you would “rise” by a certain amount while you did that. Or would you? What if the line has a negative slope? …which means that as you move in the positive X direction, the Y value decreases instead of increases? That would be like going downhill instead of uphill and the word “rise” seems less fitting in that case. Because of that, we’ll just use (delta Y) over (delta X) in this video. But it’s certainly not wrong to use the terms “rise” and “run”, especially if it help you remember the formula. But if you do, just remember that the “rise” can also be negative. Now that you’ve seen the equation for slope, let’s see the equation for calculating the distance between two points on the line. Distance equals the square root of [(delta X) squared plus (delta Y) squared]. Hmmm… does this equation remind you of anything you may have seen before? It certainly reminds me of something I’ve seen before! You stole my theorem! Ah…Um… Hi there Pythagoras. I… I didn’t steal your theorem. I’m just borrowing it so I can calculate some stuff. Don’t worry… I’ll give you credit for it. Oh…well… I guess as long as you give me credit. Do you remember how The Pythagorean Theorem tells us the relationship between the three sides of any right triangle? That means that if you know the lengths of two of the sides, you can calculate the length of the third side. Well, we just saw that if you turn a line segment into a right triangle, you can calculate the lengths of two sides which would be (delta X) and (delta Y), right? So we can just plug those values into The Pythagorean Theorem. You’re used to seeing that theorem in this form: ‘a squared’ + ‘b squared’=‘c squared’. Since ‘a’ and ‘b’ are the lengths of the horizontal and vertical sides, we can plug in (delta X) and (delta Y) instead. And instead of ‘c squared’ for the hypotenuse side, let’s use ‘d squared’ because the length of the hypotenuse equals the “distance” between our two points. To solve for ‘d’ (or distance), we take the square root of both sides and we get: d=the square root of [(delta X) squared + (delta Y) squared]. This special form of The Pythagorean Theorem is usually called “The Distance Formula” in Algebra because you can use it to find the distance between any two points on the coordinate plane. And just like with the Slope Equation, you’ll often see it in expanded form where the (delta X) and (delta Y) are written out as the subtractions X2 − X1 and Y2 − Y1. Alright, now that we have our equations for slope and distance, let’s see them in action. Suppose we’re given two coordinates and we’re asked to find the slope of the line they form, and the distance between those points on the coordinate plane. The first thing we need to do is name the coordinates since they aren’t named already. And even though it’s not really necessary, if you’re like me, you might want to draw a little sketch of the problem to help you visualize what’s going on. The equations for slope and distance both use (delta X) and (delta Y) so let’s calculate those values first. (delta X)=X2 − X1, and in this problem, X2=4 and X1=-2. So (delta X) equals 4 minus -2 which is just 6. Next we calculate (delta Y). (delta Y)=Y2 − Y1 and in this problems, Y2=3 and Y1=0. That means (delta Y)=3 minus 0 which is just 3. Great, now we have our (delta X) and (delta Y) values and you can confirm that we got them correct by looking at our graph. The length of the (delta X) side is 6 units, and the length of the (delta Y) side is 3 units. Now let’s plug those delta values into the equations for slope and distance. Slope=(delta Y) over (delta X). We just found that (delta Y)=3 and (delta X)=6, so our slope equals 3 over 6, which simplifies to one-half or 0.5 That was easy, now let’s plug those deltas into our Distance Equation to see how far apart those points are. Doing that tells us that the distance=the square root of (‘6 squared’ + ‘3 squared’). ‘6 squared’=36 and ‘3 squared’=9. 36 + 9=45, so the distance between the points would be the square root of 45. You could also simplify the answer to ‘3 root 5’ or use a calculator to convert it to a decimal which would be 6.708 (rounded to 3 places). That wasn’t so bad, was it? But let’s try one more example to make sure you’ve got it. Again, we’re given two points, but this time they’re already shown on the coordinate plane for us. We’ll label this one that’s farthest to the left, Point 1 and the other one Point 2. That means this point’s coordinates will be X1 and Y1 and this point’s coordinates will be X2 and Y2. First, we plug those coordinate values in to the delta equations. (delta X)=X2 − X1 which is 1 minus -3. So (delta X)=4 (delta Y)=Y2 − Y1 which is -2 minus 5. So (delta Y)=-7 Again, we see that those delta values agree with the graph and the right triangle formed by the line between the two points. Or do they? One of our deltas is negative, but a length can’t be negative can it? Well no, but remember that deltas are really a difference between coordinate values, so they can be negative. The lengths of the triangle’s sides are really the Absolute Values of the deltas. But the signs of the deltas are important because they help us get the correct slope, since a slope can be positive or negative. Now that we have our deltas, let’s plug them into our equation for slope. Slope=(delta Y) over (delta X) which would be -7 over 4. We could just leave the slope like that as an improper fraction. But we could also convert it into mixed number form, or get the decimal value with a calculator, which is -1.75 Now that we’ve found the slope, let’s find the distance between the points by plugging the deltas we already calculated into that equation. That gives us distance=the square root of (‘4 squared’ + ‘negative 7 squared’) ‘4 squared’=16 and ‘negative 7 squared’=49. That means the distance is the square root of (16 + 49) or the square root of 65. That root can’t be simplified, but we can use a calculator to convert it to a decimal if we want to, giving us a distance of 8.062 (rounded to three decimal places). Alright, so now you know how to calculate the slope of a line if you know any two points along that line. You also know how to calculate the distance between any two points on a line using the so called Distance Formula. Also known as The Pythagorean Theorem, invented by ME, Pythagoras. Ah yes… thank you for reminding us Pythagoras. Remember that the key to success in math is to practice, so be sure to try some slope and distance problems on your own. As always, thanks for watching Math Antics and I’ll see ya next time. Learn more at www.mathantics.com

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